The temperature of an open room of volume 30 | vedclass

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(D) Given: Initial temperature $T_i = 17 + 273 = 290 \ K$. Final temperature $T_f = 27 + 273 = 300 \ K$. Atmospheric pressure $P = 1 	imes 10^5 \ Pa$

Vedclass · Accepted Answer

$-2.5 	imes 10^{25}$

Vedclass · Answer

(D) Given: Initial temperature $T_i = 17 + 273 = 290 \ K$. Final temperature $T_f = 27 + 273 = 300 \ K$. Atmospheric pressure $P = 1 	imes 10^5 \ Pa$. Volume of the room $V = 30 \ m^3$. The number of molecules $N$ is given by the ideal gas law $PV = N k_B T$,where $k_B$ is the Boltzmann constant $(k_B = 1.38 	imes 10^{-23} \ J/K)$. Thus,$N = \frac{PV}{k_B T}$. Initial number of molecules $N_i = \frac{PV}{k_B T_i}$. Final number of molecules $N_f = \frac{PV}{k_B T_f}$. The change in the number of molecules is $N_f - N_i = \frac{PV}{k_B} \left( \frac{1}{T_f} - \frac{1}{T_i} ight)$. Substituting the values: $N_f - N_i = \frac{1 	imes 10^5 	imes 30}{1.38 	imes 10^{-23}} \left( \frac{1}{300} - \frac{1}{290} ight)$. $N_f - N_i = \frac{30 	imes 10^5}{1.38 	imes 10^{-23}} \left( \frac{290 - 300}{300 	imes 290} ight)$. $N_f - N_i = \frac{30 	imes 10^5}{1.38 	imes 10^{-23}} \left( \frac{-10}{87000} ight) \approx -2.5 	imes 10^{25}$.

The temperature of an open room of volume $30\ m^3$ increases from $17^\circ C$ to $27^\circ C$ due to sunshine. The atmospheric pressure in the room remains $1 \times 10^5\ Pa$. If $n_i$ and $n_f$ are the number of molecules in the room before and after heating,then $n_f - n_i$ will be:

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